Respuesta :
Answer:
D).75.101
Step-by-step explanation:
Previous concepts
The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}[/tex]
Where O represent the observed values and E the expected values.
State the null and alternative hypothesis
Null hypothesis: All ages have crash rates proportional to their driving rates
Alternative hypothesis: Not All ages have crash rates proportional to their driving rates
The observed values are given by the table given:
Age <26 26-45 46-65 >45
Drivers 66 39 25 30
Calculate the expected values
In order to calculate the expected values we can use the rates given by the problem:
The age distribution of drivers for the given categories is :
18% for the under 26 group, E1 = 160*0.18=28.8
39% for the 26-45 group, E2=160*0.39=62.4
31% for the 45-65 group, E3=160*0.31=49.6
12% for the group over 65. E4=160*0.12=19.2
Calculate the statistic
[tex]\chi^2 =\frac{(66-28.8)^2}{28.8}+\frac{(39-62.4)^2}{62.4}+\frac{(25-49.6)^2}{49.6}+\frac{(30-19.2)^2}{19.2}[/tex]
[tex]\chi^2 =75.101[/tex]
Calculate the critical value
First we need to calculate the degrees of freedom given by:
[tex] df= k=1=4-1= 3[/tex], where k represent the total number of categories, for tis case k=4
We can use a confidence level for example 95%, and the significance would be [tex]\alpha=1-0.95=0.05[/tex] and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be [tex]\chi^2_{crit}=7.815[/tex]
We can calculate also the p value:
[tex]p_v =P(\chi^2_{3}>75.101)=0[/tex]
And we got the same decision reject the null hypothesis at 5% of significance.
